Example Question - calculate distance
Here are examples of questions we've helped users solve.
Calculating Distance between Ships Using Trigonometry
Para resolver la pregunta en la imagen, debemos encontrar la distancia \( d \) entre los barcos A y B, utilizando trigonometría. Según la información proporcionada y el dibujo, parece que debemos aplicar la ley de los cosenos. La ley de los cosenos relaciona los lados de un triángulo con el coseno de uno de sus ángulos. La fórmula es: \[ c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \] donde \( c \) es el lado opuesto al ángulo \( C \), \( a \) y \( b \) son los otros lados del triángulo, y \( \cos(C) \) es el coseno del ángulo \( C \). En nuestro caso, tenemos un triángulo con lados de \( 3 \) km y \( 1.5 \) km, y un ángulo de \( 42 \) grados entre ellos. Queremos hallar \( d \), el lado opuesto al ángulo de \( 42 \) grados. Aplicando la ley de los cosenos: \[ d^2 = 3^2 + 1.5^2 - 2(3)(1.5) \cdot \cos(42^\circ) \] Calculamos \( \cos(42^\circ) \) y sustituimos los valores: \[ d^2 = 9 + 2.25 - 2(3)(1.5) \cdot \cos(42^\circ) \] \[ d^2 = 11.25 - 9 \cdot \cos(42^\circ) \] Ahora necesitamos calcular \( \cos(42^\circ) \), que aproximadamente es \( 0.743 \), y sustituirlo en la ecuación: \[ d^2 = 11.25 - 9(0.743) \] \[ d^2 = 11.25 - 6.687 \] \[ d^2 = 4.563 \] Tomando la raíz cuadrada de ambos lados para obtener \( d \): \[ d = \sqrt{4.563} \] \[ d \approx 2.136 km \] Entonces, la distancia \( d \) entre los barcos A y B es aproximadamente \( 2.136 \) kilómetros.
Calculating Distance Between Two Points on a Coordinate Plane
The question is asking you to calculate the distance between two points on a coordinate plane, which essentially is asking you to find the length of the hypotenuse of a right triangle formed with these two points as its endpoints. The points provided are (3, 7) and (8, 5). To determine the distance between these two points, you can use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Where (x1, y1) and (x2, y2) are the coordinates of the two points. Substitute the coordinates into the formula: \[ d = \sqrt{(8 - 3)^2 + (5 - 7)^2} \] \[ d = \sqrt{(5)^2 + (-2)^2} \] \[ d = \sqrt{25 + 4} \] \[ d = \sqrt{29} \] So, the distance between the two points to the nearest tenth is approximately \(\sqrt{29}\), which is about 5.4 when rounded to the nearest tenth.
Finding the Distance between Two Points
To find the distance between the two points (9, 1) and (3, 10), you can use the distance formula, which is derived from the Pythagorean theorem: \[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Here, \( (x_1, y_1) = (9, 1) \) and \( (x_2, y_2) = (3, 10) \). Plugging in the values: \[ \text{Distance} = \sqrt{(3 - 9)^2 + (10 - 1)^2} \] \[ \text{Distance} = \sqrt{(-6)^2 + (9)^2} \] \[ \text{Distance} = \sqrt{36 + 81} \] \[ \text{Distance} = \sqrt{117} \] Now, simplify this to get the final answer: \[ \text{Distance} = \sqrt{117} \approx 10.82 \] The question asks to round decimals to the nearest tenth: \[ \text{Distance} \approx 10.8 \text{ units} \]
Solving for Initial Velocity and Distance with an Accelerated Motion Formula
The given formula is: \[ s = ut + \frac{1}{2}at^2 \] To solve part (a), we want to make \( u \) the subject of the formula. Here's how we do it: First, let's rearrange the formula to solve for \( u \) by subtracting \( \frac{1}{2}at^2 \) from both sides of the equation: \[ s - \frac{1}{2}at^2 = ut \] Next, we divide both sides of the equation by \( t \) to solve for \( u \): \[ \frac{s - \frac{1}{2}at^2}{t} = u \] So the formula with \( u \) as the subject is: \[ u = \frac{s - \frac{1}{2}at^2}{t} \] For part (b), we need to calculate \( s \) when \( a = 100 \text{ m/s}^2 \), \( u = 2 \text{ m/s} \), and \( t = 5 \text{ s} \). We can plug the given values into the original formula: \[ s = ut + \frac{1}{2}at^2 \] \[ s = (2 \text{ m/s})(5 \text{ s}) + \frac{1}{2}(100 \text{ m/s}^2)(5 \text{ s})^2 \] \[ s = 10 \text{ m} + \frac{1}{2}(100 \text{ m/s}^2)(25 \text{ s}^2) \] \[ s = 10 \text{ m} + 50(25 \text{ m}) \] \[ s = 10 \text{ m} + 1250 \text{ m} \] \[ s = 1260 \text{ m} \] Thus, the distance \( s \) is 1260 meters.
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